poj 3620 Avoid The Lakes【简单dfs】

Avoid The Lakes

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6795		Accepted: 3622

Description

Farmer John's farm was flooded in the most recent storm, a fact only aggravated by the information that his cows are deathly afraid of water. His insurance agency will only repay him, however, an amount depending on the size of the largest "lake" on his farm.

The farm is represented as a rectangular grid with N (1 ≤ N ≤ 100) rows and M (1 ≤ M ≤ 100) columns. Each cell in the grid is either dry or submerged, and exactly K(1 ≤ K ≤ N × M) of the cells are submerged. As one would expect, a lake has a central cell to which other cells connect by sharing a long edge (not a corner). Any cell that shares a long edge with the central cell or shares a long edge with any connected cell becomes a connected cell and is part of the lake.

Input

* Line 1: Three space-separated integers: N, M, and K
* Lines 2..K+1: Line i+1 describes one submerged location with two space separated integers that are its row and column: R and C

Output

* Line 1: The number of cells that the largest lake contains.　

Sample Input

3 4 5
3 2
2 2
3 1
2 3
1 1

Sample Output

4

/*
题意：给出你湖的坐标，让你求最大的湖的大小（湖的大小是从这个点的四个方向中有一个方向
仍然为湖，即小湖可以组成大湖）
题解：先将给出的坐标储存下来并标记，其余的点记为0，
跟hdu1241做法一样，只不过此题只需要搜索四个方向 
*/
#include<stdio.h>
#include<string.h>
#define MAX 110
#define maxn(x,y)(x>y?x:y)
int map[MAX][MAX];
int tot,max,n,m;
void dfs(int x,int y)
{
    int i,j;
    int move[4][2]={0,1,0,-1,1,0,-1,0};
    if(x>0&&x<=n&&y>0&&y<=m&&map[x][y]==1)
    {
        tot++;//记录这个湖的大小 
        map[x][y]=0;
        for(i=0;i<4;i++)//四个方向查找 
        {
            dfs(x+move[i][0],y+move[i][1]);
        }
    }
}
int main()
{
    int k,r,c,t,i,j;
    memset(map,0,sizeof(map));
    scanf("%d%d%d",&n,&m,&k);
    for(i=1;i<=k;i++)
    {
        scanf("%d%d",&r,&c);
        map[r][c]=1;
    }
    max=0;
    for(i=1;i<=n;i++)
    {
        for(j=1;j<=m;j++)
        {
            if(map[i][j]==1)
            {
                tot=0;
                dfs(i,j);//搜索一遍之后tot的值就是当前搜索的湖的大小 
                max=maxn(max,tot);//找到最大的湖 
            }
        }
    }
    printf("%d
",max);
    return 0;
}